Question

10.2.8 Your answer is partially correct. Try again. A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film, and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is Ij = 1.14 and si = 0.11, while for the 20-mil film, the data yield 12 = 1.07 and s2 = 0.09. Note that an increase in film speed would lower the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use a = 0.10 and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the P-value for this test? Round your answer to three decimal places (e.g. 98.765). The data 0.013 support the claim that reducing the film thickness increases the mean speed of the film. The P-value is (b) Find a 95% confidence interval on the difference in the two means that can be used to test the claim in part (a). Round your answers to four decimal places (e.g. 98.7654). 0.0085 0.1315 KM1-42

Answer 1

(a)

Let the population mean length of 25 mil film be $\mu_1$ and that of 20 mil film be

112

Here we are to test

Ho: Mi-p2=0 against H1: Mi-u2 < 0

The given data is summarized as

	Sample 1	Sample 2
Sample size	n1=8	n2=8
Sample mean	$\bar{x}_1$=1.14	$\bar{x}_2$=1.07
Sample SD	s1=0.11	s2=0.09

The test statistic is obtained as

11 - 12 t 2 2 + ni n2

1.14 – 1.07 t= 0.07 0.05024937811 1.393052066 - -1.393 0.112 8 0.092 8

The test statistic follows t distribution with df 14

The p-value is obtained as 0.092673 i.e. 0.093

As the p-value is less than 0.1, we reject the null hypothesis at 10% level of significance.

i.e. Support the claim that reducing the film's thickness increases the mean length of the film.

(b)

The 95% confidence interval is given by

$((\bar{x}_1-\bar{x}_2)\pm \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\;\;*t_{0.025,n_1+n_2-2})$

0.112 0.092 * i.e. ((1.14 – 1.07) + * 10.025,14) < 8 00

$i.e. \;\;(0.07\pm \sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}\;\;*2.144787)$

$i.e. \;\;(0.07\pm 0.1077742129)$

$i.e. \;\;-0.0378\leq \mu_1-\mu_2 \leq 0.1778$

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